A factory has 3 machines that produce the same item. Machines 1 and 2 each produce 30% of the total output, while machines 3 produce 40%. It is known that 6% of Machine 1's output is defective, while machine 2 produces 5% defective items and machine 3 produces 4% defective items. If a randomly selected item is defective what is the probability it would be produced by machine 1?
look at 1000 items
M1 produces 300 with 18 defects
M2 produces 300 with 15 defects
M3 produces 400 with 16 defects
the probability that a defect is from M1:
18 / (18 + 15 + 16)
Ans 0.3673
To find the probability that a randomly selected item is defective and produced by machine 1, we need to use conditional probability.
Conditional probability states that the probability of an event A happening given that event B has already occurred is equal to the probability of both events happening (A and B) divided by the probability of event B happening.
In this case, event A is selecting a defective item, and event B is selecting an item produced by machine 1. We can write this as P(A|B).
Let's break down the information given:
- Machines 1 and 2 each produce 30% of the total output, and machine 3 produces 40%. This means that the probability of selecting an item produced by machine 1 is 30%.
- Machine 1 produces 6% defective items, machine 2 produces 5% defective items, and machine 3 produces 4% defective items.
To calculate P(A and B), the probability of selecting a defective item produced by machine 1, we multiply the probability of selecting a defective item (A) by the probability of selecting an item produced by machine 1 (B).
P(A and B) = P(A) * P(B)
P(A and B) = 0.06 * 0.30
P(A and B) = 0.018
To calculate P(B), the probability of selecting an item produced by machine 1, we don't need any additional information since it is already given as 30%.
P(B) = 0.30
To find P(A|B), the probability of selecting a defective item given that it was produced by machine 1, we divide P(A and B) by P(B), using the formula for conditional probability:
P(A|B) = P(A and B) / P(B)
P(A|B) = 0.018 / 0.30
P(A|B) ≈ 0.06
Therefore, the probability that a randomly selected defective item is produced by machine 1 is approximately 0.06, or 6%.